Question Number 140139 by mathsuji last updated on 04/May/21 Answered by mr W last updated on 04/May/21 $${say}\:{radius}\:{of}\:{curcumcircle}\:{is}\:{r} \\ $$$$\Sigma\mathrm{sin}^{−\mathrm{1}} \frac{{a}}{\mathrm{2}{r}}=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{6}}{\mathrm{2}{r}}+\mathrm{sin}\:\frac{\mathrm{3}}{\mathrm{2}{r}}+\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{11}}}{\mathrm{2}{r}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{6}}{\mathrm{2}{r}}+\mathrm{sin}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\mathrm{2}{r}}=\pi…
Question Number 74599 by mind is power last updated on 27/Nov/19 $$\mathrm{Hello},\mathrm{verry}\:\mathrm{Nice}\:\mathrm{day}\: \\ $$$$\mathrm{let}\:\mathrm{U}_{\mathrm{n}} =\mathrm{E}\left(\left(\frac{\mathrm{3}+\sqrt{\mathrm{17}}}{\mathrm{2}}\right)^{\mathrm{n}} \right),\mathrm{n}\in\mathbb{N}^{\ast} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{U}_{\mathrm{n}} \equiv\mathrm{n}\left(\mathrm{2}\right) \\ $$ Terms of Service Privacy…
Question Number 9061 by tawakalitu last updated on 16/Nov/16 Commented by tawakalitu last updated on 16/Nov/16 $$\mathrm{please}\:\mathrm{help}. \\ $$ Answered by mrW last updated on…
Question Number 140129 by mathdanisur last updated on 04/May/21 $${x};{y}\in\mathbb{R}^{+} \:;\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2} \\ $$$${proof}:\:\mathrm{3}−{xy}\geqslant\left({x}+{y}\right)\sqrt{{xy}}+\left({x}−{y}\right)^{\mathrm{2}} \geqslant\mathrm{2}{xy} \\ $$ Answered by mr W last updated on…
Question Number 74587 by lalitchand last updated on 27/Nov/19 $$\mathrm{If}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{n}\:\mathrm{arithmetic}\:\mathrm{means}\:\mathrm{between}\: \\ $$$$\mathrm{two}\:\mathrm{number}\:\mathrm{is}\:\mathrm{20}.\mathrm{if}\:\mathrm{last}\:\mathrm{mean}\:\mathrm{is}\:\mathrm{double} \\ $$$$\mathrm{of}\:\mathrm{1st}\:\mathrm{mean}\:\mathrm{and}\:\mathrm{one}\:\mathrm{is}\:\mathrm{three}\:\mathrm{times}\:\mathrm{of} \\ $$$$\mathrm{another}\:\mathrm{number}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{numbers}. \\ $$$$ \\ $$ Terms of Service Privacy Policy…
Question Number 140113 by EnterUsername last updated on 04/May/21 $$\mathrm{Let}\:{z}_{\mathrm{1}} =\mathrm{1}+{i},\:{z}_{\mathrm{2}} =−\mathrm{1}−{i}\:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers} \\ $$$$\mathrm{such}\:\mathrm{that}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{form}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}. \\ $$$$\mathrm{Then}\:{z}_{\mathrm{3}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{A}\right)\:\sqrt{\mathrm{3}}\left(\mathrm{1}+{i}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\sqrt{\mathrm{3}}\left(\mathrm{1}−{i}\right) \\ $$$$\left(\mathrm{C}\right)\:\sqrt{\mathrm{3}}\left({i}−\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\sqrt{\mathrm{3}}\left(−\mathrm{1}−{i}\right)…
Question Number 140114 by EnterUsername last updated on 04/May/21 $$\mathrm{If}\:\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{cos}\:\gamma=\mathrm{0}=\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma,\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{cos}\left(\mathrm{2}\alpha\right)+\mathrm{cos}\left(\mathrm{2}\beta\right)+\mathrm{cos}\left(\mathrm{2}\gamma\right)=\mathrm{0} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{sin}\left(\mathrm{3}\alpha\right)+\mathrm{sin}\left(\mathrm{3}\beta\right)+\mathrm{sin}\left(\mathrm{3}\gamma\right)=\mathrm{3sin}\left(\alpha+\beta+\gamma\right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{cos}\left(\mathrm{3}\alpha\right)+\mathrm{cos}\left(\mathrm{3}\beta\right)+\mathrm{cos}\left(\mathrm{3}\gamma\right)=\mathrm{3cos}\left(\alpha+\beta+\gamma\right) \\ $$$$\left(\mathrm{D}\right)\:\mathrm{sin}\left(\mathrm{2}\alpha\right)+\mathrm{sin}\left(\mathrm{2}\beta\right)+\mathrm{sin}\left(\mathrm{2}\gamma\right)=\mathrm{0} \\ $$ Terms of Service Privacy Policy…
Question Number 140104 by EnterUsername last updated on 04/May/21 $$\mathrm{Let}\:{z}=\mathrm{1}+\mathrm{cos}\left(\mathrm{10}\pi/\mathrm{9}\right)+{i}\mathrm{sin}\left(\mathrm{10}\pi/\mathrm{9}\right).\:\mathrm{Then} \\ $$$$\left(\mathrm{A}\right)\:\mid{z}\mid=\mathrm{2cos}\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{arg}\:{z}=\frac{\mathrm{8}\pi}{\mathrm{9}} \\ $$$$\left(\mathrm{C}\right)\:\mid{z}\mid=\mathrm{2cos}\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{arg}\:{z}=\frac{\mathrm{5}\pi}{\mathrm{9}} \\ $$ Commented by EnterUsername last updated on 04/May/21 $$\mathrm{One}\:\mathrm{or}\:\mathrm{more}\:\mathrm{answers}\:\mathrm{may}\:\mathrm{be}\:\mathrm{correct}. \\…
Question Number 140107 by EnterUsername last updated on 04/May/21 $$\mathrm{If}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$$\mathrm{described}\:\mathrm{in}\:\mathrm{counterclock}\:\mathrm{sense}\:\mathrm{and}\:{w}\neq\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cube}\:\mathrm{root} \\ $$$$\mathrm{of}\:\mathrm{unity},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:{z}_{\mathrm{1}} −{z}_{\mathrm{3}} =\left({z}_{\mathrm{3}} −{z}_{\mathrm{2}} \right){w} \\ $$$$\left(\mathrm{B}\right)\:{z}_{\mathrm{1}}…
Question Number 140102 by mr W last updated on 04/May/21 $${if}\:{a}_{{n}} −\mathrm{5}{a}_{{n}−\mathrm{1}} +\mathrm{6}{a}_{{n}−\mathrm{2}} =\mathrm{1}\:{and}\: \\ $$$${a}_{\mathrm{0}} =\mathrm{1},\:{a}_{\mathrm{1}} =\mathrm{2}. \\ $$$${find}\:{a}_{{n}} \:{in}\:{terms}\:{of}\:{n}. \\ $$ Commented by…